nyoj322 sort 归并排序，树状数组

Sort

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

描述

You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need.
For example, 1 2 3 5 4, we only need one operation : swap 5 and 4.

输入

The input consists of T number of test cases.(<0T<1000) Each case consists of two lines: the first line contains a positive integer n (n <= 1000); the next line contains a permutation of the n integers from 1 to n.

输出

For each case, output the minimum times need to sort it in ascending order on a single line.

样例输入

2
3
1 2 3
4
4 3 2 1

样例输出

0
6

如果按冒泡排序这些O(n^2)肯定会超时，所以需要找一种更快的方法 --------归并排序。

归并排序是建立在归并操作上的一种有效的排序算法。该算法是采用分治法的一个非常典型的应用。将已有序的子序列合并，得到完全有序的序列；即先使每个子序列有序，再使子序列段间有序。若将两个有序表合并成一个有序表，称为2-路归并。这题还可以用树状数组来做

法一：用归并排序做

#include<stdio.h>
int a[1000001],b[1000001]; /*合并排序结果先保存到b中*/
int merge(int a[],int low,int mid,int high)
{
    int i=low,j=mid+1,k=low;
    int count=0;/*计数器*/ 
    while((i<=mid)&&(j<=high))/*部分合并*/
    {
        if(a[i]<=a[j])
        {
            b[k++]=a[i++];
        }
        else
        {
            b[k++]=a[j++];
            count+=(mid-i+1);
        }
    }
    while(i<=mid)/*转储剩余部分*/
    {
        b[k++]=a[i++];
    }
    while(j<=high)
    {
        b[k++]=a[j++];
        
    }
    for(i=low;i<=high;++i)/*把b中的值复制给a*/
    {
        a[i]=b[i];
    }
    return count;
}
int sort(int a[],int low,int high)
{
    int x,y,z;
    int mid=(high+low)/2;
    int i=low,j=mid+1;
    if(low>=high)
    {
        return 0;
    }
    x=sort(a,low,mid);
    y=sort(a,mid+1,high);
    z=merge(a,low,mid,high);
    return (x+y+z);
}
int main()
{
    int ncases,n,i;
    scanf("%d",&ncases);
    while(ncases--)
    {
        scanf("%d",&n);
        for(i=0;i<=n-1;i++)
        {
            scanf("%d",&a[i]);
        }
        printf("%d
",sort(a,0,n-1));
    }
    return 0;
}

 法二：用树状数组

 不知道什么是树状数组的  就先看看树状数组吧。。链接：http://dongxicheng.org/structure/binary_indexed_tree/

#include<stdio.h>
#include<string.h>
int num[1004],n;
int lowbit(int x)
{
    return x&(-x);
}
void add(int x)
//更新含有x的数组个数
{
    while(x<=n)
    {
        num[x]++;
        x+=lowbit(x);
    }
}
int sum(int x)
//向下统计小于x的个数
{
    int total=0;
    while(x>0)
    {
        total+=num[x];
        x-=lowbit(x);
    }
    return total;
}
int main()
{
    int x,cases;
    scanf("%d",&cases);
    while(cases--)
    {
        scanf( "%d",&n);
        memset( num,0,sizeof( num ));
        int ss = 0;
        for( int i = 0; i < n; ++i )
        {
            scanf( "%d",&x);
            add(x);
            ss += (i-sum( x - 1 ));
        }
        printf( "%d
",ss );
    }
    return 0;
}